The value of $\sqrt{11}$ lies between which two consecutive integers ? Integers that appear in order when counting, for example 2 and 3.
Solution: Consider the perfect squares near $11$ . [ What are perfect squares? Perfect squares are integers which can be obtained by squaring an integer. The first 13 perfect squares are: $ 1,4,9,16,25,36,49,64,81,100,121,144,169$ $9$ is the nearest perfect square less than $11$ $16$ is the nearest perfect square more than $11$ So, we know $9 < 11 < 16$ So, $\sqrt{9} < \sqrt{11} < \sqrt{16}$ So $\sqrt{11}$ is between $3$ and $4$.